Chapter 14Gas-Vapor Mixtures and Air-Conditioning Study Guide in PowerPoint to accompany Thermodynamics: An Engineering Approach, 7th edition by Yunus A. Çengel and Michael A. Boles We will be concerned with the mixture of dry air and water vapor. This mixture is often called atmospheric air. The temperature of the atmospheric air in air-conditioning applications ranges from about –10 to about 50oC. Under these conditions, we treat air as an ideal gas with constant specific heats. Taking Cpa = 1.005 kJ/kgK, the enthalpy of the dry air is given by (assuming the reference state to be 0oC where the reference enthalpy is taken to be 0 kJ/kga) The assumption that the water vapor is an ideal gas is valid when the mixture temperature is below 50oC. This means that the saturation pressure of the water vapor in the air-vapor mixture is below 12.3 kPa. For these conditions, the enthalpy of the water vapor is approximated by hv(T) = hg at mixture temperature T. The following T-s diagram for water illustrates the ideal-gas behavior at low vapor pressures. See Figure A-9 for the actual T-s diagram. 2 For more information and animations illustrating this topic visit the Animation Library developed by Professor S. Bhattacharjee, San Diego State University, at this link. test.sdsu.edu/testhome/vtAnimations/index.html The saturated vapor value of the enthalpy is a function of temperature and can be expressed as Note: For the dry air-water vapor mixture, the partial pressure of the water vapor in the mixture is less that its saturation pressure at the temperature. Pv Psat @Tmix 3 the partial pressure of the water vapor can never be greater than its saturation pressure corresponding to the temperature of the mixture. P v When the mixture pressure is increased while keeping the mixture temperature constant. the vapor partial pressure increases up to the vapor saturation pressure at the mixture temperature and condensation begins. Therefore. Assume that the water vapor is initially superheated. See if you can sketch the process on the P-v diagram relative to the saturation lines for the water alone given below.Consider increasing the total pressure of an air-water vapor mixture while the temperature of the mixture is held constant. 4 . Definitions Dew Point.95oC. When the mixture is cooled to a temperature equal to the saturation temperature for the water-vapor partial pressure.491 kPa. condensation begins. Steam 375 325 275 T [C] 225 175 1. then the dew point temperature of that mixture is 12. Tdp The dew point is the temperature at which vapor condenses or solidifies when cooled at constant pressure.491 kPa 125 75 TDP 25 -25 0 2 4 6 s [kJ/kg-K] 8 10 12 5 . When an atmospheric air-vapor mixture is cooled at constant pressure such that the partial pressure of the water vapor is 1. Consider cooling an air-water vapor mixture while the mixture total pressure is held constant. Relative Humidity. Steam 125 T [C] 75 Tm o Vapor State Pg = 3. Pv 1491 . kPa 6 .491 kPa -25 0 2 4 6 8 10 12 s [kJ/kg-K] Since Pg Pv . ϕ Mass of vapor in air m v Mass of in saturated air mg Pv Pg Pv and Pg are shown on the following T-s diagram for the water-vapor alone.169 kPa 25 Tdp Pv = 1. 1 or 100%.47 Pg 3169 . kPa 0. the relative humidity may be expressed as 0.Absolute humidity or specific humidity (sometimes called humidity ratio).622 Pa P Pv Using the definition of the specific humidity.622 ) Pg P Pg Volume of mixture per mass of dry air. we can show (you should try this) v V RT va a m ma Pa 7 .622 v 0. v v V m R T /P m m m m ma ma After several steps.622Pg P and (0. Mass of water vapor in air mv Mass of dry air ma PVM Pv M v v v / ( Ru T ) PaVM a / ( Ru T ) Pa M a P Pv 0. the mass flow rate of dry air is V m a v m3 / s kga 3 m / kga s Enthalpy of mixture per mass dry air. h Hm Ha Hv ma ha mv hv h ma ma ma ha hv 8 . Mass of mixture mv m ma mv ma (1 ) ma (1 ) ma Mass flow rate of dry air.So the volume of the mixture per unit mass of dry air is the specific volume of the dry air calculated at the mixture temperature and the partial pressure of the dry air. m a Based on the volume flow rate of mixture at a given state. 100 kPa.548 kPa 0.247 kPa 9 . and h of the mixture per mass of dry air. has a dew point of 21.3oC.6 or 60% Pg 4. Pv 2. humidity ratio.Example 14-1 Atmospheric air at 30oC. Find the relative humidity. (30 C )) o kga C kga kgv kJ 7171 . T) kJ kgv kJ o o 1005 . 182 .01626 (100 2.548) kPa kga h ha hv C p .622 0.2. 182 . (30 C ) 0.01626 (25013 . a T ( 25013 . kga 10 .548 kPa kgv 0. Pg = 2.1 mv .01626) kgv 0.4(2.339 kPa ) 0. 2 mv .1 ma ma 2 1 (0.936 kPa 0.936 kPa Pv 0.936) kPa kg 0.00588 0. what mass of water is added or removed per unit mass of dry air? At 20oC. 2 mv .622 The change in mass of water per mass of dry air is mv .01038 kga kgv kga 11 .339 kPa. 40 percent relative humidity.Example 14-2 If the atmospheric air in the last example is conditioned to 20oC.622 P Pv (100 0.00588 v kga w 0. Pv Pg 0. 01038 kg of water vapor is condensed for each kg of dry air.1 MPa. Example 14-3 Atmospheric air is at 25oC.1 Tsat @ Pv 13. and with 1 = 50% o s Pv .170 kPa ) 1.1 1 Pg .585 kPa Tdp . 0. T At 25 C. Sketch the water-vapor states relative to the saturation lines on the following T-s diagram. 0.170 kPa.1 0.Or. Psat = 3.8o C 12 .5(3. 50 percent relative humidity. find the amount of water removed per mass of dry air. as the mixture changes from state 1 to state 2. If the mixture is cooled at constant pressure to 10oC. when the mixture gets cooled to T2 = 10oC < Tdp.2 = Pg. ) kPa kgv kga Therefore. the mixture is saturated.622 Pv .2 P Pv .1 0.00228 v kga 13 .228 kPa. w2 0.1 P Pv .622 1. kPa (100 15845 .00773 0.2 = 1.622 15845 .01001) kga kg 0.228) kPa kg v kg a The change in mass of water per mass of dry air is mv .622 Pv .01001 0.1. 2 mv . 2 0.228 kPa (100 1.1 ma 2 1 kgv (0. and 2 = 100%.00773 0.w1 0. Then Pv. determine the required rate of heat transfer. 0.00228 kg of water vapor is condensed for each kg of dry air. Cooling fluid In Out Insulated flow duct Atmospheric air T2 = 20oC T1 = 30oC P1 =100 kPa 1 = 80% V&1 = 17m3/min Condensate at 20oC P2 = 98 kPa 2 = 95% 14 .Or as the mixture changes from state 1 to state 2. Example 14-3 Given the inlet and exit conditions to an air conditioner shown below. What is the heat transfer to be removed per kg dry air flowing through the device? If the volume flow rate of the inlet atmospheric air is 17 m3/min. Steady-Flow Analysis Applied to Gas-Vapor Mixtures We will review the conservation of mass and conservation of energy principles as they apply to gas-vapor mixtures in the following example. Some say the water leaves at the average of 26 and 20oC. Apply the conservation of energy to the steady-flow control volume 2 2 V V Q net m i (h gz ) i W net m e (h gz ) e 2 2 inlets exits Neglecting the kinetic and potential energies and noting that the work is zero.1 1 Pg .Before we apply the steady-flow conservation of mass and energy. some water-vapor will condense. however. we need to decide if any water is condensed in the process.1 Tsat @ Pv 26.01o C So for T2 = 20oC < Tdp.8(4. Is the mixture cooled below the dew point for state 1? Pv . 20oC is adequate for our use here. we get Q net m a1ha1 m v1hv1 m a 2 ha 2 m v 2 hv 2 m l 2 hl 2 Conservation of mass for the steady-flow control volume is m m i inlets e exits 15 . Let's assume that the condensed water leaves the air conditioner at 20oC. 1.247 kPa ) 3.1 0.398 kPa Tdp . For the dry air: m a1 m a 2 m a For the water vapor: m v1 m v 2 m l 2 The mass of water that is condensed and leaves the control volume is m l 2 m v1 m v 2 m a ( 1 2 ) a . then Divide the conservation of energy equation by m Q net ha1 1hv1 ha 2 2 hv 2 ( 1 2 )hl 2 m a Q net ha 2 ha1 2 hv 2 1hv1 ( 1 2 )hl 2 m a Q net C pa (T2 T1 ) 2 hv 2 1hv1 ( 1 2 )hl 2 m a 16 . 222 kgv 0.622 Pv1 0.Now to find the 's and h's.339 kPa ) 2.398) 1 P1 Pv1 100 3.398 kg v 0.95)(2.222 kPa 0.622(3.222) 2 P2 Pv 2 98 2.02188 kg a Pv 2 2 Pg 2 (0.622 Pv 2 0.01443 kga 17 .622(2. 0. Using the steam tables.91 ) kg a kg v kg a kg v kJ kg a 18 .73 kg v kg kJ kJ (2555.6 kJ kgv hv 2 2537.01443) v (83.4 ) o kg a C kg a kgv 0.02188 0.01443 (2537.6 ) (0.02188 28.91 kJ kg v The required heat transfer per unit mass of dry air becomes Q&net C pa (T2 T1 ) 2 hv 2 1hv1 (1 2 ) hl 2 m&a 1. the h's for the water are hv1 2555.005 kg v kJ kJ o (20 30) C 0.4 kJ kgv hl 2 83. 2843 Tons 2.73 m&a kg a The mass flow rate of dry air is given by v1 RaT1 Pa1 kJ (30 273) K 3 kg a K m kPa (100 3.398) kPa kJ 0.287 m3 0.73 ) min kg a 60s kJ 9.90 kg a V1 m a v1 m3 17 min 18.The heat transfer from the atmospheric air is qout Q&net kJ 28.89 a (28.90 kg a kg kJ 1min 1kWs Q&out m&a qout 18.89 kg a m&a m3 min 0.57 Tons of refrigeration kW 19 .04 kW 0. 20 . some of the water will evaporate and the temperature of the air-vapor mixture will decrease.The Adiabatic Saturation Process Air having a relative humidity less than 100 percent flows over water contained in a well-insulated duct. Since the air has < 100 percent. the temperature of the mixture on leaving the device is known as the adiabatic saturation temperature. We assume that the total pressure is constant during the process. makeup water at the adiabatic saturation temperature is added at the same rate at which water is evaporated. we get m a1ha1 m v1hv1 m l 2 hl 2 m a 2 ha 2 m v 2 hv 2 Conservation of mass for the steady-flow control volume is m m i inlets e exits 21 . For this to be a steady-flow process.If the mixture leaving the duct is saturated and if the process is adiabatic. Apply the conservation of energy to the steady-flow control volume 2 2 V V Q net m i (h gz ) i W net m e (h gz ) e 2 2 inlets exits Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero. then ha1 1hv1 ( 2 1 )hl 2 ha 2 2 hv 2 What are the knowns and unknowns in this equation? 22 .m a1 m a 2 m a For the dry air: For the water vapor: m v1 m l 2 m v 2 The mass flow rate water that must be supplied to maintain steady-flow is. m l 2 m v 2 m v1 m a ( 2 1 ) Divide the conservation of energy equation by m a . 622 P1 Pv1 1 P1 Pv1 0. the relative humidity at state 1 is Pv1 1 Pg1 23 . C pa (T2 T1 ) 2 h fg 2 (hg1 h f 2 ) Pv1 1 0.622 1 Then.Solving for 1 ha 2 ha1 2 (hv 2 hl 2 ) 1 (hv1 hl 2 ) Since 1 is also defined by We can solve for Pv1. Example 14-4 For the adiabatic saturation process shown below. and enthalpy of the atmospheric air per mass of dry air at state 1. 24 . determine the relative humidity. humidity ratio (specific humidity). 0 kJ kg v From the above analysis 1 C pa (T2 T1 ) 2 h fg 2 (hg1 h f 2 ) 1.Using the steam tables: hf 2 kJ 67.7 kg v h fg 2 2463.7 67.0 ) o kg a kg v kg a C kJ (2544.005 kg v kJ kJ o 16 24 C 0.00822 kg v kg a 25 .0115 (2463.2) kg v 0.2 kg v kJ hv1 2544. 005 kg v kJ kJ o (24 C ) 0.004 kPa The enthalpy of the mixture at state 1 is h1 ha1 1hv1 C paT1 1hv1 1.3 kPa 0.We can solve for Pv1.3 kPa Pv1 Then the relative humidity at state 1 is 1 Pv1 Pv1 Pg1 Psat @ 24o C 1.433 or 43.00822(100kPa) 0.00822 1.3% 3.04 kJ kg a 26 .7 kg a o C kg a kg v 45. 1 P1 0.00822 2544.622 1 0.622 0. temperature. resulting in a lower temperature to be registered by the thermometer. The dryer the atmospheric air. there will be little difference between the wet-bulb and dry-bulb temperatures. 27 . When the relative humidity of the air is near 100 percent. The other thermometer reads the dry-bulb. As the psychrometer is slung through the air. One thermometer is fitted with a wet gauze and reads the wetbulb temperature. The psychrometer is composed of two thermometers mounted on a sling. the state of atmospheric air is specified by determining the wetbulb and dry-bulb temperatures. water vaporizes from the wet gauze. or ordinary. These temperatures are measured by using a device called a psychrometer. The wet-bulb and dry-bulb temperatures and the atmospheric pressure uniquely determine the state of the atmospheric air. the lower the wet-bulb temperature will be.Wet-Bulb and Dry-Bulb Temperatures In normal practice. The wet-bulb temperature is approximately equal to the adiabatic saturation temperature. fixed. total air-vapor pressure.The Psychrometric Chart For a given. the properties of the mixture are given in graphical form on a psychrometric chart. The air-conditioning processes: 28 . 29 . 008 kga kga kJ h 46 kga m3 v 0.Example 14-5 Determine the relative humidity.853 kga NOTE: THE ENTHALPY READ FROM THE PSYCHROMETRIC CHART IS THE TOTAL ENTHALPY OF THE AIR-VAPOR MIXTURE PER UNIT MASS OF DRY AIR. From the psychrometric chart read 44% gv kgv 8.0 0. h= H/ma = ha + ωhv 30 . humidity ratio (specific humidity). enthalpy of the atmospheric air per mass of dry air. and atmospheric pressure is 100 kPa. the wet-bulb temperature is 16oC. and the specific volume of the mixture per mass of dry air at a state where the dry-bulb temperature is 24oC. determine the required heat transfer rate in the heating section and the required steam temperature in the humidification section when the steam pressure is 1 MPa.Example 14-6 For the air-conditioning system shown below in which atmospheric air is first heated and then humidified with a steam spray. 31 . 793 m^3/kga 15 20 25 30 35 40 T [C] Apply conservation of mass and conservation of energy for steady-flow to process 1-2.025 0.015 0.0049 kgv/kga 2 0 5 10 v1 =0.8 0.020 h3 =48 kJ/kga 0. Conservation of mass for the steady-flow control volume is m m i inlets e exits 32 .2 0C -5 0.010 0.3 [kPa] 0.The psychrometric diagram is Psychrometric Diagram 0.4 1 = 2 =0.0091kgv/kga 0.040 Humidity Ratio 0.030 30 C 0.035 0.050 0.005 0.000 -10 20 C h2 =37 kJ/kga h1 =17 kJ/kga1 3 10 C 3 =0.045 Pressure = 101.6 0. 2 1 Neglecting the kinetic and potential energies and noting that the work is zero.For the dry air m a1 m a 2 m a For the water vapor (note: no water is added or condensed during simple heating) m v1 m v 2 Thus. and letting the enthalpy of the mixture per unit mass of air h be defined as h ha hv we obtain E in E out Q in m a h1 m a h2 Q in m a (h2 h1 ) 33 . and T2 = 24oC: The mass flow rate of dry air is given by V1 m a v1 34 . a and h's using the psychrometric chart. 1 = 90%. Now to find the m At T1 = 5oC. The psychrometric chart gives 35 . state 3.793 kga The required heat transfer rate for the heating section is kga kJ 1kWs Qin 1261 . 3 m min 60s s 0. At the exit. T3 = 25oC and 3 = 45%.m3 60 kga 1 min kga min m a 75. (37 17) s kga kJ 25.66 1261 . List some ways in which this amount of heat can be supplied.22 kW This is the required heat transfer to the atmospheric air. (0.0089 0. Conservation of mass for the steady-flow control volume is m m i inlets For the dry air e exits m a 2 m a 3 m a For the water vapor (note: water is added during the humidification process) m v 2 m s m v 3 m s m v 3 m v 2 m s m a ( 3 2 ) kga kgv 1261 .0049) s kga kgv 0.Apply conservation of mass and conservation of energy to process 2-3.00504 s 36 . Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero. m a ( 3 2 )hs m a (h3 h2 ) h3 h2 hs 3 2 37 . the conservation of energy yields E in E out m a h2 m s hs m a h3 m s hs m a (h3 h2 ) Solving for the enthalpy of the steam. 0049) kg a kJ 2750 kg v At Ps = 1 MPa and hs = 2750 kJ/kgv. evaporative cooling.985. adiabatic mixing of airstreams. See the text for applications involving cooling with dehumidification.hs kJ (48 37) kg a kg v (0.0089 0. 38 . and wet cooling towers.88oC and the quality xs = 0. Ts = 179.